# Inequality 34( Pham Kim Hung)

Problem:

If $\displaystyle a,b,c$ are positive real numbers, prove that $\displaystyle \frac{a^2+b^2+c^2}{ab+bc+ca}+\frac{8abc}{(a+b)(b+c)(c+a)}\geq 2$.

# Inequality 32(Vasile Cirtoaje)

The following inequality and its solution is dedicated to my beloved teacher, Christos Patilas…

Problem:

If $\displaystyle a,b,c,d$ are positive real numbers, with $\displaystyle abcd=1$, prove that

$\displaystyle \frac{1}{(1+a)(1+a^2)}+\frac{1}{(1+b)(1+b^2)}+\frac{1}{(1+c)(1+c^2)}+\frac{1}{(1+d)(1+d^2)}\geq 1$.

Solution:

Without loss of generality assume that $\displaystyle a\geq b\geq c\geq d$. Then from Chebyshev’s inequality we have

$\displaystyle \frac{1}{(1+a)(1+a^2)}+\frac{1}{(1+b)(1+b^2)}+\frac{1}{(1+c)(1+c^2)}\geq \frac{1}{3}\left(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\right)\left(\frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}\right)$.

Lemma (Vasile Cirtoaje):

If $\displaystyle a\geq b\geq c\geq d$ and $\displaystyle abcd=1$ then it holds that

$\displaystyle \frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\geq \frac{3}{1+\sqrt[3]{abc}}$.

Proof of the Lemma:

We know that $\displaystyle \frac{1}{1+a}+\frac{1}{1+b}\geq \frac{2}{1+\sqrt{ab}}$. So, it suffices to prove that $\displaystyle \frac{1}{1+c}+\frac{2}{1+\sqrt{ab}}\geq \frac{3}{1+\sqrt[3]{abc}}$.

Let us denote by $\displaystyle x=\sqrt{ab},y=\sqrt[3]{abc}\Longrightarrow c=\frac{y^3}{x^2}$. Substituting them to the above inequality we get that

$\displaystyle \frac{1}{1+c}+\frac{2}{1+\sqrt{ab}}-\frac{3}{1+\sqrt[3]{abc}}=\frac{x^2}{x^2+y^3}+\frac{2}{1+x}-\frac{3}{1+y}$,

which reduces to the inequality

$\displaystyle \frac{(x-y)^2\left[2y^2-y+x(y-2)\right]}{(1+x)(1+y)(x^2+y^3)}$,

which is obvious since $\displaystyle 2y^2-y+(y-2)x\geq 2y^2-y+(y-2)y^3=y(y-1)(y^2-y+1)\geq 0$.

Back to our inequality now, from the above lemma we deduce that:

$\displaystyle \frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\geq \frac{3}{1+\sqrt[3]{abc}}\wedge \frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}\geq \frac{3}{1+\sqrt[3]{a^2b^2c^2}}$.

For convenience denote by $\displaystyle k$ the $\displaystyle \sqrt[3]{abc}$. Therefore we have that

$\displaystyle \frac{1}{(1+a)(1+a^2)}+\frac{1}{(1+b)(1+b^2)}+\frac{1}{(1+c)(1+c^2)}\geq \frac{3}{(1+k)(1+k^2)}$.

Thus it remains to prove that

$\displaystyle \frac{3}{(1+k)(1+k^2)}+\frac{1}{(1+d)(1+d^2)}\geq 1$.

But $\displaystyle abcd=1$. So, the last fraction is of the form $\displaystyle \frac{1}{\left(1+\frac{1}{k^3}\right)\left(1+\frac{1}{k^6}\right)}$.

After that we get

$\displaystyle \frac{1}{\left(1+\frac{1}{k^3}\right)\left(1+\frac{1}{k^6}\right)}+\frac{3}{(1+k)(1+k^2)}\geq 1$.

Conclusion follows from the obvious inequalityÂ  $\displaystyle \frac{(k-1)^2(2k^4+k^3+k+2)}{(k^3+1)(k^6+1)}\geq 0$

# Inequality 31(Generalization, Vo Quoc Ba Can)

Problem:

Let $\displaystyle a,b,c,d$ be positive real numbers. Prove that for $\displaystyle k\in \left[\frac{1}{2},2\right]$, the following inequality holds:

$\displaystyle \left(\frac{a+kb}{a+b+c}\right)^2+\left(\frac{b+kc}{b+c+d}\right)^2+\left(\frac{c+kd}{c+d+a}\right)^2+\left(\frac{d+ka}{d+a+b}\right)^2\geq \frac{4}{9}(k+1)^2$.

Solution:

Wrong soluion :(. Can, i wait for your solution on that problem ðŸ˜€

# Inequality 30(APMO 2005)

Problem:

Let $\displaystyle x,y,z$ be positive real numbers such that $\displaystyle xyz=8$. Prove that

$\displaystyle \frac{x^2}{\sqrt{(x^3+1)(y^3+1)}}+\frac{y^2}{\sqrt{(y^3+1)(z^3+1)}}+\frac{z^2}{\sqrt{(z^3+1)(x^3+1)}}\geq\frac{4}{3}$.

Solution:

From the AM-GM ineuqality we know that

$\displaystyle \frac{1}{\sqrt{x^3+1}}=\frac{1}{\sqrt{(x+1)(x^2-x+1)}}\geq \frac{2}{(x+1)+(x^2-x+1)}=\frac{2}{x^2+2}$.

Doing that cyclic over the $\displaystyle 3$ fractions we get that

$\displaystyle \sum_{cyc}\frac{x^2}{\sqrt{(x^3+1)(y^3+1)}}\geq \frac{4x^2}{(x^2+2)(y^2+2)}$.

So, it suffices to prove that

$\displaystyle \frac{4x^2}{(x^2+2)(y^2+2)}\geq \frac{4}{3}$, or $\displaystyle 3\sum_{cyc}x^2(z^2+2)\geq \prod_{cyc}(x^2+2)$.

After expanding, the inequality is equivalent to

$\displaystyle 2\sum_{cyc}x^2+\sum_{cyc}x^2y^2\geq 72$.

But the last relation holds due to the AM-GM inequality and from the hypothesis, since $\displaystyle \sum_{cyc}x^2y^2\geq 3\sqrt[3]{8^4}=48\wedge 2\sum_{cyc}x^2\geq 6\sqrt[3]{8^2}=24$.

Adding up these $\displaystyle 2$ relations we get the desired result, Q.E.D.

# Inequality 29(George Basdekis)

Problem:

If $\displaystyle x,y,z,r>0$ prove that

$\displaystyle \frac{2}{3}r\cdot\left(\sum_{cyc}x^2\right)^{2}+r^{2}\sum_{cyc}x^4\sum_{cyc}x^2y^2+1\geq 4r\sum_{cyc}x^3y$.

We know that

$\displaystyle 4r\sum_{cyc}x^3y\leq \left(r\sum_{cyc}x^3y+1\right)^2=r^2\left(\sum_{cyc}x^3y\right)^2+2r\sum_{cyc}x^3y+1$.

So, we only need to prove that

$\displaystyle \frac{2}{3}r\cdot\left(\sum_{cyc}x^2\right)^{2}+r^{2}\sum_{cyc}x^4\sum_{cyc}x^2y^2+1\geq r^2\left(\sum_{cyc}x^3y\right)^2+2r\sum_{cyc}x^3y+1$,

or

$\displaystyle \frac{2}{3}\cdot r\left(\sum_{cyc}x^2\right)^2+r^2\sum_{cyc}x^4\sum_{cyc}x^2y^2\geq r^2\left(\sum_{cyc}x^3y\right)^2+2r\sum_{cyc}x^3y$.

But from Cauchy-Schwarz Inequality we deduce that

$\displaystyle r^2\sum_{cyc}x^4\sum_{cyc}x^2y^2\geq r^2\left(\sum_{cyc}x^3y\right)^2$.

So ,we only need to prove now that

$\displaystyle \frac{2}{3}\cdot r\left(\sum_{cyc}x^2\right)^2\geq 2r\sum_{cyc}x^3y$,

which is obviously true from Cirtoaje’s Inequality, Q.E.D.

# Inequality 28(Russian Winter Olympiad 2006)

Problem:

Let $\displaystyle a,b,c,d$ be positive real numbers such that $\displaystyle a+b+c+d=4$. Prove that

$\displaystyle \frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+d^2}+\frac{d}{1+a^2}\geq 2$.

Solution:

Let us make the Cauchy reverse technique and apply AM-GM inequality. Then we have that:

• $\displaystyle \frac{a}{1+b^2}=a-\frac{ab^2}{1+b^2}\leq a-\frac{ab}{2}$,
• $\displaystyle \frac{b}{1+c^2}=b-\frac{bc^2}{1+b^2}\leq b-\frac{bc}{2}$,
• $\displaystyle \frac{c}{1+d^2}=c-\frac{cd^2}{1+d^2}\leq c-\frac{cd}{2}$,
• $\displaystyle \frac{d}{1+a^2}=d-\frac{da^2}{1+a^2}\leq d-\frac{da}{2}$.

So, it is enough to prove that

$\displaystyle \sum_{cyc}a-\frac{1}{2}\sum_{cyc}ab\geq 2\Longrightarrow 4\geq ab+bc+cd+da$.

Multiplying by $\displaystyle 4$ we get that $\displaystyle (a+b+c+d)^2\geq 4(ab+bc+cd+da)$, which reduces to the obvious inequality

$\displaystyle (a-b+c-d)^2\geq 0$, Q.E.D.

# Inequality 27(Vo Quoc Ba Can)

Problem:

Let $\displaystyle a,b,c$ be positive real numbers. Prove that

$\displaystyle \frac{a^{b+c}}{(b+c)^2}+\frac{b^{c+a}}{(c+a)^2}+\frac{c^{a+b}}{(a+b)^2}\geq \frac{3}{4}$

Solution:

It holds for every positive number $\displaystyle x$ that

$\displaystyle \frac{1}{x^2}\geq \frac{1}{x^{2x}}$.

Back to our inequality now, rewrite it as follows:

$\displaystyle \frac{a^{b+c}}{(b+c)^2}+\frac{b^{c+a}}{(c+a)^2}+\frac{c^{a+b}}{(a+b)^2}\geq \frac{3}{4}\Longrightarrow \sum_{cyc}\frac{4a^{b+c}}{(b+c)^2}\geq 3\Longrightarrow \sum_{cyc}\frac{a^{b+c}}{\left(\frac{b+c}{2}\right)^{2}}\geq 3$.

Using the above result we have that

$\displaystyle \sum_{cyc}\frac{a^{b+c}}{\left(\frac{b+c}{2}\right)^{2}}\geq \sum_{cyc}\left(\frac{a}{\frac{b+c}{2}}\right)^{b+c}=\sum_{cyc}\left(\frac{2a}{b+c}\right)^{b+c}$.

Let us now make the final step for the proof of our inequality. From Bernulli’s inequality we get that

$\displaystyle \sum_{cyc}\left(\frac{2a}{b+c}\right)^{b+c}=\sum_{cyc}\left[\left(\frac{2a}{b+c}-1\right)+1\right]^{b+c}\geq \sum_{cyc}(1+2a-b-c)=3$, Q.E.D.